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Original scientific paper
https://doi.org/10.3336/gm.55.1.04

Perfect powers in an alternating sum of consecutive cubes

Pranabesh Das   ORCID icon orcid.org/0000-0001-9119-5402 ; Pure Mathematics, University of Waterloo, 200 University Avenue West, Waterloo, Ontario, Canada
Pallab Kanti Dey ; Stat-Math Unit, Indian Statistical Institute, 7, S. J. S. Sansanwal Marg, New Delhi, Delhi - 110016, India
Bibekananda Maji   ORCID icon orcid.org/0000-0003-2155-2480 ; Department of Mathematics, Indian Institute of Technology Indore, Simrol, Indore, Madhya Pradesh - 453552, India
Sudhansu Sekhar Rout ; Institute of Mathematics and Applications, Andharua, Bhubaneswar, Odisha - 751029, India

Fulltext: english, pdf (182 KB) pages 37-53 downloads: 145* cite
APA 6th Edition
Das, P., Dey, P.K., Maji, B. & Rout, S.S. (2020). Perfect powers in an alternating sum of consecutive cubes. Glasnik matematički, 55 (1), 37-53. https://doi.org/10.3336/gm.55.1.04
MLA 8th Edition
Das, Pranabesh, et al. "Perfect powers in an alternating sum of consecutive cubes." Glasnik matematički, vol. 55, no. 1, 2020, pp. 37-53. https://doi.org/10.3336/gm.55.1.04. Accessed 19 Oct. 2021.
Chicago 17th Edition
Das, Pranabesh, Pallab Kanti Dey, Bibekananda Maji and Sudhansu Sekhar Rout. "Perfect powers in an alternating sum of consecutive cubes." Glasnik matematički 55, no. 1 (2020): 37-53. https://doi.org/10.3336/gm.55.1.04
Harvard
Das, P., et al. (2020). 'Perfect powers in an alternating sum of consecutive cubes', Glasnik matematički, 55(1), pp. 37-53. https://doi.org/10.3336/gm.55.1.04
Vancouver
Das P, Dey PK, Maji B, Rout SS. Perfect powers in an alternating sum of consecutive cubes. Glasnik matematički [Internet]. 2020 [cited 2021 October 19];55(1):37-53. https://doi.org/10.3336/gm.55.1.04
IEEE
P. Das, P.K. Dey, B. Maji and S.S. Rout, "Perfect powers in an alternating sum of consecutive cubes", Glasnik matematički, vol.55, no. 1, pp. 37-53, 2020. [Online]. https://doi.org/10.3336/gm.55.1.04

Abstracts
In this paper, we consider the problem about finding out perfect powers in an alternating sum of consecutive cubes. More precisely, we completely solve the Diophantine equation (x+1)3 - (x+2)3 + ∙∙∙ - (x + 2d)3 + (x + 2d + 1)3 = zp, where p is prime and x,d,z are integers with 1 ≤ d ≤ 50.

Keywords
Diophantine equation; Galois representation; Frey curve; modularity; level lowering; linear forms in logarithms

Hrčak ID: 239041

URI
https://hrcak.srce.hr/239041

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